Problem: You have found the following ages (in years) of all 6 snakes at your local zoo: $ 2,\enspace 17,\enspace 17,\enspace 11,\enspace 2,\enspace 1$ What is the average age of the snakes at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 snakes at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 17 + 17 + 11 + 2 + 1}{{6}} = {8.3\text{ years old}} $ Find the squared deviations from the mean for each snake. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-6.3$ years $39.69$ years $^2$ $17$ years $8.7$ years $75.69$ years $^2$ $17$ years $8.7$ years $75.69$ years $^2$ $11$ years $2.7$ years $7.29$ years $^2$ $2$ years $-6.3$ years $39.69$ years $^2$ $1$ year $-7.3$ years $53.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{39.69} + {75.69} + {75.69} + {7.29} + {39.69} + {53.29}} {{6}} $ $ {\sigma^2} = \dfrac{{291.34}}{{6}} = {48.56\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{48.56\text{ years}^2}} = {7\text{ years}} $ The average snake at the zoo is 8.3 years old. There is a standard deviation of 7 years.